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Programming in C MCQ - Find the Output

26 Find the output of the following C code (assume #include <stdio.h> and that the code is inside int main(void)):

int i = 1, sum = 0;
for (; i < 20;) {
    sum += i;
    i = i * 2 + 1;
}
printf("%d %d", sum, i);

The body sees i = 1, 3, 7, 15; sum = 26 and the terminating i = 31.

27 Find the output of the following C code (assume #include <stdio.h> and that the code is inside int main(void)):

int i, j = 10, count = 0;
for (i = 1; (j -= 2) > i; i += 2)
    count += j;
printf("%d %d", count, j);

The successful tests assign j=8 and j=6. The final condition assigns j=4, which is not greater than i=5.

28 Find the output of the following C code (assume #include <stdio.h> and that the code is inside int main(void)):

int i, j, x = 0;
for (i = 1; i <= 3; i++) {
    for (j = 1; j <= i; j++) {
        x += j;
        if (x % 3 == 0)
            break;
    }
}
printf("%d", x);

The outer passes leave x = 1, 4, and 10. No accumulated value reaches a multiple of three at the inner break checks.

29 Find the output of the following C code (assume #include <stdio.h> and that the code is inside int main(void)):

int i = 1, count = 0;
for (; i++ < 7;) {
    if (i % 2 == 0)
        count += i;
}
printf("%d %d", count, i);

The body sees i=2 through 7. Even values 2,4,6 contribute 12; the failed condition increments i to 8.

30 Find the output of the following C code (assume #include <stdio.h> and that the code is inside int main(void)):

int i, j, count = 0;
for (i = 1; i <= 4; i++) {
    for (j = 1; j <= i; j++) {
        if (i == j)
            continue;
        count += i - j;
    }
}
printf("%d", count);

The outer passes contribute 0, 1, 3, and 6, totaling 10.

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Question:

Find the output of the following C code (assume #include <stdio.h> and that the code is inside int main(void)):int i, j, sum = 0; for (i = 1;…

A. 24
B. 30
C. 32
D. 40

Correct answer: C

Explanation:

Only odd i+j pairs contribute: 6 for i=1, 8 for i=2, and 18 for i=3, totaling 32.

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Question

Find the output of the following C code (assume #include <stdio.h> and that the code is inside int main(void)):

int i, j, sum = 0;
for (i = 1; i <= 3; i++) {
    for (j = 1; j <= 4; j++) {
        if ((i + j) % 2 == 0)
            continue;
        sum += i * j;
    }
}
printf("%d", sum);

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